If the function $y = f(x)$ satisfies the differential equation $(x^3 + 1)dy = x(1 - 3xy)dx$ and $f(0) = 0$,then $\mathop {\lim }\limits_{x \to 0} \frac{x^2}{f(x)}$ is equal to

Integrating factor of $(x+2 y^3) \frac{d y}{d x}=y^2$ is

By multiplying with $e^{\int P dx}$ on both sides of the equation $\frac{dy}{dx} + P(x)y = Q(x)$,the left side of the equation takes the form $\frac{d}{dx}(y f(x))$,then $f(x) =$

Let $f$ be a non-negative function defined on $\left[0, \frac{\pi}{2}\right]$. If $\int_0^x \left(f^{\prime}(t)-\sin 2t\right) dt = \int_x^0 f(t) \tan t dt$ and $f(0)=1$, then find $\int_0^{\frac{\pi}{2}} f(x) dx$.

Let $f(x)$ be differentiable on the interval $(0, \infty)$ such that $f(1)=1$,and $\lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$ for each $x>0$. Then $f(x)$ is

The solution of the differential equation $y' = y \tan x - 2 \sin x$ is